The LCM of 40, 78 and another number, x, is 7800. Which of the following can be the value of x?

Correct option is B

Given:

LCM of 40, 78, x = 7800

We are to find which of the given options could be a possible value of x 

Solution:

Prime factorizations:

40 = $$2^3 \times 5$$​​

78 = $$2 \times 3 \times 13$$​​

LCM(40, 78) = $$2^3 \times 3 \times 5 \times 13$$​= 1560

Now,

LCM(40, 78, x) = 7800 

From the option;

Option A: 180

180 = $$2^2 \times 3^2 \times 5$$​​

Combined with 40 and 78:

Max powers: $$2^3, 3^2, 5, 13$$​​

LCM = $$2^3 \times 3^2 \times 5 \times 13$$​= 4680 Not equal to 7800

Option B: 200

200 = $$2^3 \times 5^2$$​​

Now take max powers with 40 and 78: $$2^3, 3, 5^2, 13$$​​

LCM = $$2^3 \times 3 \times 5^2 \times 13$$​ = 7800 equal to 7800

Option C: 127

127 is a prime number, not related to 2, 3, 5, or 13

LCM with 40 and 78 would be 1560 $$\times$$​ 127 = 198120 Not equal to 7800

Option D: 152

152 = $$2^3 \times 19$$​​

New prime = 19 → LCM becomes

1560 × 19 = 29640 Not equal to 7800

Thus, correct option is B) 200